題目連結：
POJ 3243 Clever Y
題意：
跟POJ 2417 Discrete Logging類似，只不過gcd(a,p)!=1.
分析；
初始化cnt=0(消因子輪數),d=1(消掉的gcd乘積).
令tmp=gcd(a,p) 當tmp!=1時,修改變數值:b/=tmp(先判斷b是否是tmp的倍數),p/=tmp, d=a/tmp∗d
通過若干輪消掉a,p的因子使得最終gcd(a,p)=1.這時再呼叫普通的BSGS得到解為res，則最終答案是res+cnt。
但是這樣求得的解是≥cnt的，需要先判斷下是否有<cnt的解。考慮cnt的最大值。因為每次消去的最小因子是2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
 

using namespace std;
typedef long long ll;

const ll MOD = 100007;
ll hs[MOD + 100], id[MOD + 100];

ll find(ll x)
{
    ll t = x % MOD;
    while(hs[t] != x && hs[t] != -1) t = (t + 1) % MOD;
    return t;
}

void insert(ll x, ll ii)
{
    ll pos = find(x);
    if(hs[pos] == -1){
        hs[pos] = x;
        id[pos] = ii;
    }
}

ll get(ll x)
{
    ll pos = find(x);
    return
 
 hs[pos] == x ? id[pos] : -1;
}

ll ex_gcd(ll a, ll b, ll& x, ll& y)
{
    if(b == 0) {
        x = 1, y = 0;
        return a;
    }
    ll d = ex_gcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

ll inv(ll a, ll p)
{
    ll d, x, y;
    d = ex_gcd(a, p, x, y);
    return d == 1 ? (x % p + p) % p : -1;
}

ll BSGS(ll a, ll b, ll p)
{
    memset(hs, -1, sizeof(hs));
    memset(id, -1, sizeof(id));
    ll m = (ll)ceil(sqrt(p + 0.5));
    ll tmp = 1;
    for(ll i = 0; i < m; i++) {
        insert(tmp, i);
        tmp = tmp * a % p;
    }
    ll base = inv(tmp, p);
    ll res = b, z;
    for(ll i = 0; i < m; i++) {
        if((z = get(res)) != -1) return i * m + z;
        res = res * base % p;
    }
    return -1;
}

ll gcd(ll x, ll y)
{
    return y == 0 ? x : gcd(y, x % y);
}

ll solve(ll a, ll b, ll p)
{
    ll tmp = 1;
    for(int i = 0; i <= 50; ++ i) {
        if(tmp == b) return i;
        tmp = tmp * a % p;
    }
    ll cnt = 0, d = 1 % p;
    while((tmp = gcd(a, p)) != 1) {
        if(b % tmp) return -1;
        b /= tmp;
        p /= tmp;
        d = a / tmp * d % p;
        cnt++;
    }
    b = b * inv(d, p) % p;
    ll ans = BSGS(a, b, p);
    if(ans == -1) return -1;
    else return ans + cnt;
}

int main()
{
    ll a, b, p;
    while(~scanf("%lld%lld%lld", &a, &p, &b) && (a || p || b)){
        ll ans = solve(a, b % p, p);
        if(ans == -1) printf("No Solution\n");
        else printf("%lld\n", ans);
    }
    return 0;
}